Answer to Question #240570 in Classical Mechanics for smith

Question

[#446] Going up, going down (part E)

We how have:

Net force on m

m

Net force on M

M

ma = T-mg

ma=T−mg

Ma=Mg-T

Ma=Mg−T

Using these results from part D, find a very simple expression for the acceleration a

a in terms of g

g.

Acceleration = _______.

Answer 1

We have:

$$a=g-\frac{T}{M},\\\space\\ T=m(a+g).\\\space\\ a=g-\frac{m(a+g)}{M}=g-\frac{ma}{M}-\frac{mg}{M},\\\space\\ a=g\frac{M-m}{M+m}.$$